An earthquake \"P\" wave passes across a boundary in rock where its velocity inc

Question

An earthquake "P" wave passes across a boundary in rock where its velocity increases from 6.0 km/s to 8.0 km/s. If it strikes this boundary at 30Degree, what is the angle of refraction? 22 38 42 46 None of the Above During a lightning storm, the sound of thunder reaches your car 11 seconds after you sec the lightning. How far away is the lightning strike 3.0 km 3.8 km 5.1 km 31 km None of the Above At a busy street comer, the sound level is 130 dB. What is the intensity of sound in W/m^2? 1 times 10^1 1 times 10Degree 1 times 10^minus1 1 times 10^minus2 None of the Above What is the new sound level when 10 students arc talking given that each talks at the 100 dB level? 105 dB 110 dB 112 dB 115 dB 150 dB The sound level measured 20 in from a jet plane is 120 dB. What is the sound level at 300 m? (Ignore reflections from the ground.) 130 dB 120 dB 96 dB 70 dB 1 dB

Explanation / Answer

17)

It's just like light refraction; Snell's law applies.

Speed ratio v2/v1 = N1/N2 = sin(theta2)/sin(theta1)

v2/v1 = sin(theta incidence) / sin(theta refraction)

By putting the values

8/6 = sin(30) / sin(theta refraction)

angle of refraction = 22.02431284 degree.

18)

As we know that speed of light = 3 * 10^8 m/s

So in 11 secs it travels = 3 * 11 * 10^8 m = 3.3 * 10^9m

19)

As we know that

Decibals are basically the ratio the intesity of a sound level to a constant intensity level (which is the lowest intesity detectable by the human ear) with a base 10 logarithm thrown in. Equation form looks like this:

dB = 10log I/Io; where I = given intensity and Io = constant (1*10^-12 W/m^2).

So to find the first one we would go:

130 = 10log I/1*10^-12, solving for "I" would get you:

I = (1*10^-12)*10^13 = 10 W/m^2

20)

Sound intensity I = Io×10^(LI/10) W/m² = 10^12×10^(140/10) W/m² = 100 W/m².

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